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I'm lacking some vital understanding about quaternions and algebras in general. If we first define $V=\{a+bi+cj+dk|a,b,c,d\in\mathbb{R}\}$. Then we define scalar multiplication, vector multiplication, addition e.t.c. But as I understand it, aren't $1,i,j,k$ meant to be vectors, meaning that $bi$ for example has to already be defined even for the set to be defined. Very confused! As usual thanks for any replies!

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  • $\begingroup$ You don't need to consider $i, j, k$ as vectors. Just consider them to be symbols which satisfy certain relations such as $i^2 = j^2 = k^2 = -1$. $\endgroup$
    – Old John
    Jan 10 '14 at 21:22
  • $\begingroup$ That doesn't cut it for me though. I like knowing what I'm working with! $\endgroup$
    – Lammey
    Jan 10 '14 at 21:42
  • $\begingroup$ FWIW, the formal symbol route can be made rigorous easily. Interpret ``$a + bi + cj + dk$'' as meaning the function from a set of four (distinct) elements $\{1, i, j, k\}$ to $\mathbb{R}$ given by $1 \mapsto a, i \mapsto b, j \mapsto c, k \mapsto d$. Such functions have a natural $\mathbb{R}$-vector space structure (pointwise operations). In this language $i$ is an abuse of notation which means the corresponding indicator function. This is just a (more coordinate free...) repackaging of the usual tuple approach. $\endgroup$ Mar 7 '16 at 21:07
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What you write for $V$ is not a precise definition (strictly speaking). One defines $\mathbb{H} := \mathbb{R}^4$ as a $\mathbb{R}$-vector space, so that quaternions are $4$-tuples $(a,b,c,d)$ of real numbers $a,b,c,d \in \mathbb{R}$. If we define $1 := (1,0,0,0)$, $i := (0,1,0,0)$, $j := (0,0,1,0)$ and $k := (0,0,0,1)$, then by construction we have $(a,b,c,d) = a1 + bi + cj + dk$. After that, one defines the multipliciation in such a way that the relations $i^2=j^2=k^2=-1$, $ij=k$ are satisfied. This is a bit cumbersome, especially when one wants to check that $\mathbb{H}$ is in fact a skew field. A more elegant way is to define $\mathbb{H}$ as the subalgebra $\left\{\begin{pmatrix} z & w \\ -\overline{w} & \overline{z} \end{pmatrix} : z,w \in \mathbb{C}\right\}$ of $M_2(\mathbb{C})$.

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  • $\begingroup$ So when I write 1, i, j, k, am I referring to (1,0,0,0) e.t.c.? or am I saying 1, i, j, k are vectors whose form I can't know i.e. they may be matrices, or anything isomorphic to $\mathbf{H}$, and the only thing I know about them is their relationships with each other? Thanks for the reply by the way. $\endgroup$
    – Lammey
    Jan 10 '14 at 21:55
  • $\begingroup$ I would say this depends on your general point of view of mathematics. As a category theorist, I agree that only the relations between $1,i,j,k$ matter. But since you have asked and were confused by the formal definition of the quaternions, I have given the most explicit definition. There are also other definitions (apart from the two in my answer, I can imagine three further ones), which all turn out to be isomorphic. You may choose your favorite one. Also compare this with the construction of complex numbers; there are several ways to make $\mathbb{C}=\{a+bi : a,b \in \mathbb{R}\}$ precise. $\endgroup$ Jan 10 '14 at 21:59
  • $\begingroup$ I had the same trouble with the complex numbers, I ended up seeing them as pairs of real numbers, and concluding that the square root of -1 doesn't really exist. I guess my problem is that I when I'm reading through proofs/textbooks in the future and someone refers to the set $\mathbb{H}$, I want to be able to know what they're talking about. I.e. are the elements of $\mathbb{H}$ specific vectors i.e. tuples, or are they more general? It seems like the symbol $\mathbb{H}$ refers to anything isomorphic to $\mathbb{H}$. And yet if I say, what is the expression a+bi+cj+dk to someone, they'll $\endgroup$
    – Lammey
    Jan 10 '14 at 22:23
  • $\begingroup$ say "it's a quaternion, duh", but then if I ask what are "1,i,j,k" will they say "also quaternions"? That sounds like a circular definition to me! $\endgroup$
    – Lammey
    Jan 10 '14 at 22:24
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    $\begingroup$ Yes it is a bit circular when one isn't used to it. Therefore, I suggest that you choose the definitions I gave in my answer (until you find another definition which works better for you). It is really simple, quaternions "are" just 4-tuples of real numbers, with the usual scalar multiplication and addition, and with a kind of complicated multiplication, but which may be reduced to the simple rules $i^2=j^2=k^2=-1$, $ij=k$. By the way, $\mathbb{C}$ definitely has a square root of $-1$, in fact two such square roots $(\pm i)$. $\endgroup$ Jan 10 '14 at 22:32
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In the quaternion algebra $\;\Bbb H\;,\;\;1,i,j,k\;$ are just abstract elements, and then there are defined operations and relations: $\;i^2=j^2=k^2=-1\;,\;\;ij=k\;$ , etc.

I don't understand what's the problem with seeing these algebra's (or any other's, of course) elements as "vectors": as part of a four dimensional real vector space, all the elements are vectors, and thus you can see $\;bi\;,\;\;b\in\Bbb R\;$ the scalar product of the vector $\;i\;$ with the scalar $\;b\;$ , or as part of a ring, the product of the two ring elements $\;b\,,\,i\;$. Both these products, formally different, are the same since an algebra behaves well in this respect.

In short: $\;\Bbb H\;$ can be seen as a four dimensional real vector space which is also a non-commutative ring (in fact, a division algebra).

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  • $\begingroup$ But if you say $1,i,j,j$ are all just vectors, and so are all linear combinations of them, it still doesn't tell me what these vectors are! I can't define a quaternion to be a combination of other quaternions! $\endgroup$
    – Lammey
    Jan 10 '14 at 21:57
  • $\begingroup$ Yes @JamesMachin...so? That wasn't your question, as far as I see it. Anyway, according to the definitions that you surely have, it is very easy to show that, referring to the vector space $\;\Bbb H_{\Bbb R}\;$ , the set $\;\{1,i,j,k\}\;$ is linearly independent and a generator set of $\;\Bbb H\;$ (i.e., a basis) ... $\endgroup$
    – DonAntonio
    Jan 10 '14 at 22:00
  • $\begingroup$ Yes I could, but my question is basically, what is "1,i,j,k". If you define a quaternion as q=a+bi+cj+dk, then you can't say "i is a quaternion. What are the elements of $\mathbb{H}$ then? like could you write one down explicitly? $\endgroup$
    – Lammey
    Jan 10 '14 at 22:27
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Mull over this: when you define polynomials over $x$ in terms of $x$, do you have trouble accepting $x$ alone as a polynomial?

If it helps, you could just think of $1,i,j,k$ as symbols, with no special properties other than being distinct. Then you define quaternions by

  • declaring the underlying set to be formal expresions of the form $a+bi+cj+dk$, whose nicknames are "quaternions."
  • declaring all the addition multiplication rules you need to make the implied addition and multiplication mean something official (like $1\cdot x=x$, $ij=-ji$, $ai+bi=(a+b)i$, distributivity, etc.)

When both of these steps are done, the four original symbols are represented in the formal set: e.g. $0\cdot 1+1i+0k+0j$ is the same thing as $i$, and henceforth would be a "quaternion."


Both polynomial rings and quaternions can be viewed as building rings out of generators and relations.

I'm guessing you're familiar with polynomial rings in two variables, like $F[x,y]$. When we build those there is an assumption that $xy=yx$... but do we really need to assume that? No! The notation $F\langle x,y\rangle$ is often used to denote the polynomial ring in "noncommuting indeterminates," and it has more elements in the sense that $xy$ and $yx$ are different things, $xyx$ and $x^2y$ are different, and so on.

(And really, there's nothing stopping us from defining an even more general type of polynomial ring where the indeterinates are nonassociating, and $(xy)z\neq x(yz)$ and so on. But we have to stop digressing for the sake of discussion :) )

But $F\langle x,y \rangle$ and $F[x,y]$ are definitely related: its easy to see that $F[x,y]\cong F\langle x,y \rangle/(xy-yx)$, that is, the polynomial ring is a quotient of the noncommuting polynomial ring. The thing in the ideal we're modding out by is referred to as a relation because it specifies some extra property that we are enforcing in the quotient algebra.

Another example of this would be $\Bbb R[i]/(i^2+1)\cong\Bbb C$: you see, the symbol $i$ (nobody said anything about an imaginary number yet) is forced to have the property $i^2=-1$ in the quotient ring. Thus $i$ in the quotient ring becomes a complex number, along with all the other elements.

By now you're noticing that if this makes sense for complex numbers, then why not quaternions? Indeed, if you take $\Bbb H\cong \Bbb R\langle i,j,k\rangle/(i^2+1,j^2+1,k^2+1,ijk+1)$ since those are all the relations we need to bend the noncommuting polynomial ring into the quaternions. (Those aren't necessarily the best or the only relations that will do the trick, but I guess they are the most familiar.)

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